Q15 of 149 Page 15

Find the value of k if points (k, 3), (6, - 2) and (-3, 4) are collinear.

The three given points are A(k, 3), B(6, −2) and C(−3, 4). It is also said that they are collinear and hence the area enclosed by them should be 0.


Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)


= |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|


Given that area of ∆ABC = 0


∴ 0 = |k(-2 – 4) + 6(4 - 3) - 3(3 – (-2))|


∴ 0 = |-6k + 6 - 15|


∴ - |-6k + 9|= 0


6k + 9 = 0


∴ k =


Hence, the value of k is


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